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    <p>EDIT: I've just thought of something which could <em>certainly</em> cause a problem. Do you ever <em>modify</em> the array in <code>Matrix</code>? You <em>mustn't</em> change the relevant values within a key after adding it to a map - it will potentially cause false positives and even false negatives (you could use the same key reference you used in <code>put</code> and not find a match with <code>get</code> if you've made a change which affects the hash code, for example).</p> <hr> <p>It's not clear what code is printing the "Recognized hashCode" part... but it sounds like it's <em>assuming</em> hash codes are unique. <strong>Don't do that.</strong> They're not <em>supposed</em> to be unique. They're meant to be an indication of <em>possible</em> equality - an optimization to make finding something faster. The idea is that if you find a matching hash code, you should <em>always</em> check for "real" equality.</p> <p>You haven't said what the values in your array are, but assuming they're 0 (empty) 1 (black) and 2 (white) there are <em>far</em> more possible combinations of those than a plain ol' 32-bit hash code can represent... you've got 3<sup>64</sup> possibilities, which is rather more than 2<sup>32</sup>. Now I dare say most of those aren't really feasible Othello boards, but I'd be surprised if <code>Arrays.deepHashCode</code> attempted to optimize for exactly this case, in order to provide unique hash codes for all valid Othello boards, even if that <em>is</em> possible.</p> <p>As a side note, your <code>hashCode</code> method is rather more complicated than it needs to be. Try this:</p> <pre><code>@Override public int hashCode() { return Arrays.deepHashCode(board); } </code></pre>
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    1. POWhy does my HashMap recognize hash keys that aren't supposed to be keys yet?
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    1. USJon Skeet
    1. USJon Skeet
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    1. POWhy does my HashMap recognize hash keys that aren't supposed to be keys yet?
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    1. COTo expand the concept, the contract between equals() and hashcode() is that given o1 and o2, o1.hashcode() == o2.hashcode() if and only if o1.equals(o2) is true (equals objects have the same hashcode). The reverse is not true (same hashcode does not imply equality).
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    2. CONote: `new Long((long) i << 32 ^ i ^ m)` have the same hashCode() of `m` for all values of `n >= 0` (its a bit more complicate for negative values) i.e. ~4 billion Long values have the same hashCode(). ;)
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      1. USPeter Lawrey
    3. CO@Soronthar No, it is not "if and only if" it is merely "if". o1.equals(o2) can be false even if o1.hashCode() == o2.hashCode().
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